Question

One quarter sector is cut from a uniform circular disc of radius \(\mathrm{R}\). This sector has mass \(\mathrm{M}\). It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is... \(\\{\mathrm{A}\\}(1 / 2) \mathrm{MR}^{2}\) \(\\{\mathrm{B}\\}(1 / 4) \mathrm{MR}^{2}\) \(\\{\mathrm{C}\\}(1 / 8) \mathrm{MR}^{2}\) \\{D \(\\} \sqrt{2} \mathrm{MR}^{2}\)

Short answer

The moment of inertia of the quarter sector is \(\cfrac{1}{4} MR^2\), which corresponds to answer choice B.

Step-by-step solution

Identify the variables

We are given the radius of the circular disc, R, and the mass of the quarter sector, M.

Calculate the moment of inertia of the circular disc

We first calculate the moment of inertia of the complete uniform circular disc, which is given by the equation \(I_{disc} = \cfrac{1}{2}MR^2\), where M is the mass and R is the radius of the disc.

Calculate the mass of the original disc

We need to find the mass of the original disc to determine the mass density. Since the quarter sector has mass M, the original disc has a mass of 4M.

Calculate the mass density of the disc

We can find the mass density, ρ, of the original disc by dividing the total mass by the total area. The mass density ρ is given by: \[ρ = \cfrac{4M}{πR^2}\]

Calculate the moment of inertia of the sector

For the quarter sector, we first calculate the moment of inertia for a small mass element, dm, with respect to the axis of rotation. Then, we will integrate over the entire quarter sector. The mass element dm can be written as: \[dm = ρ \cdot dA\] where dA = r dr dθ, and θ is the angle between the outer radius and the quarter sector. So, dm = \(\frac{4M}{πR^2} rdrdθ\) Now, let's plug dm into the moment of inertia equation: \[dI = r^2 dm = r^2 \cdot \frac{4M}{πR^2} rdrdθ\] Next, we integrate dI to find the moment of inertia I of the quarter sector: \[I_{sector} = \int\int r^2 \cdot \frac{4M}{πR^2} rdrdθ\] For the quarter sector, we integrate from r=0 to R and θ=0 to π/2: \[I_{sector} = \cfrac{4M}{πR^2} \int_{0}^{R} r^3 dr \int_{0}^{\frac{π}{2}} dθ\]

Perform the integrations and simplify

Perform the integrations and simplify the expression: \[I_{sector} = \cfrac{4M}{πR^2} \cdot \left[\cfrac{r^4}{4}\right]_{0}^{R} \cdot \left[θ\right]_{0}^{\frac{π}{2}}\] \[I_{sector} = \cfrac{1}{4} MR^2\] The moment of inertia of the quarter sector is (1/4)MR², which corresponds to answer choice B.

Key concepts

circular disc

A circular disc is a flat, round object often found in mathematics and physics problems. Imagine a smooth, flat Frisbee or a CD, and you'll have a pretty good picture of it. In theoretical exercises, a circular disc is usually considered to be perfectly uniform and perfectly circular, which means:

• It has no irregularities or deviations in thickness or density across its surface.
• Every part of the disc is equally spaced from the center, forming a perfect circle.

This uniformity allows for straightforward calculations and applications of formulas. Understanding this concept is key when dealing with physical properties like moment of inertia because it ensures that mass distribution is even. The moment of inertia is significantly influenced by how the mass of the disc is distributed, which happens to be perfect in a theoretical circular disc.

axis of rotation

The axis of rotation refers to an imaginary line around which an object, such as our circular disc, rotates. This line can pass through the center of the disc, as in our exercise, or it can be offset elsewhere. Here are some points to remember:

• If the axis passes through the center of mass, the object rotates smoothly because the mass is evenly distributed around the axis.
• The moment of inertia, or the resistance to rotational motion, changes depending on the axis location—as mass is distributed differently relative to different axes.
• In our exercise, because the quarter sector rotates about an axis through the center of the original disc, calculations focus on radial distances from this point.

Understanding the axis of rotation helps in figuring out how an object will move and behave when it spins, whether in real life or in physics problems.

mass density

Mass density is a measure of how much mass is present in a given volume or area. In our circular disc scenario, it's crucial to understand because:

• Mass density affects how weight is spread across a surface, impacting calculations for things like moment of inertia.
• For a circular disc, mass density is uniform, meaning mass is spread evenly across the disc's area, making calculations feasible with straightforward math.

Mathematically, mass density, represented as \( \rho \), is defined as:\[ \rho = \frac{\text{Total Mass}}{\text{Total Area}} \]In this exercise, the density helps us determine how to distribute a continuous mass over the area of interest when trying to solve the moment of inertia for a portion of the disc. In practical applications, understanding density aids in designing and evaluating objects based on their strength, stability, and rotational features.

integral calculus

Integral calculus is a branch of mathematics that lets us compute areas, masses, and volumes of objects when the size varies continuously. In problems involving a circular disc, calculus helps us:

• Break down complex shapes into simpler differential elements that are easier to manage mathematically.
• Integrate these small elements over a set range to find total properties like area, mass, and, importantly, the moment of inertia.

For our exercise, integral calculus involves using a double integral to sum up the small mass elements across the quarter sector. The generic mass element can be written as:\[ dm = \rho \cdot dA \]By integrating these elements over the defined limits of the circular segment, we compute its total moment of inertia:\[ I_{\text{sector}} = \int \int r^2 dm \]This integration allows us to account for all mass contributions from tiny slices and radial segments, giving a clear understanding of the whole sector's resistance to rotation. Learning how to apply integral calculus this way is foundational for solving various physics and engineering problems.